Question

Let $f=\{(x,\frac{x^2}{1+x^2}):x\in R\}$ be a function from $R$ to $R$. Then the range of $f$ is

• A. $[0,1)$
• B. $[0,1]$
• C. $[0,2)$
• D. $Noneofthese$

Answer 1

The correct option is A $[0,1)$

Given, $f=\{(x,\frac{x^2}{1+x^2}):x\in R\}$ and $f:R\to R$

Let $y=\frac{x^2}{1+x^2}$

$\because f\left(x\right)$ or $y$ is positive for all values of $x$

and

$1+x^2>x^2$
$\Rightarrow 0\le y<1$

$\therefore$ Range of $f=\{y:y\in Randy\in [0,1)\}$.